Problem: $f(n) = -4n+6-2(h(n))$ $g(t) = -6t^{2}-t+1-f(t)$ $h(t) = t^{3}-2t^{2}$ $ h(f(1)) = {?} $
Solution: First, let's solve for the value of the inner function, $f(1)$ . Then we'll know what to plug into the outer function. $f(1) = (-4)(1)+6-2(h(1))$ To solve for the value of $f$ , we need to solve for the value of $h(1)$ $h(1) = 1^{3}-2(1^{2})$ $h(1) = -1$ That means $f(1) = (-4)(1)+6+(-2)(-1)$ $f(1) = 4$ Now we know that $f(1) = 4$ . Let's solve for $h(f(1))$ , which is $h(4)$ $h(4) = 4^{3}-2(4^{2})$ $h(4) = 32$